The graph of stopping potential ' $\mathrm{V}_{\mathrm{s}}$ ' against frequency ' $v$ ' of incident…

The graph of stopping potential ' $\mathrm{V}_{\mathrm{s}}$ ' against frequency ' $v$ ' of incident radiation is plotted for two different metals ' X ' and ' Y ' as shown in graph. ' $\phi_{\mathrm{x}}$ ' and ' $\phi_{\mathrm{y}}$ ' are work functions of ' x ' and ' $Y$ ', respectively then

1. $\phi_x=\phi_y$
2. $\phi_x \lt \phi_y$
3. $\phi_x\gt\phi_y$
4. $\phi_x=\phi_y=0$

Solution

We know $\phi=h v_0 \quad \Rightarrow \phi \propto v_0$
Also, $\begin{aligned} & v_0 \lt v_0^{\prime} \\ & \therefore \quad \phi_x \lt \phi_y ..(From graph) \end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 2)

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